Question: Anjali's plane had been flying through calm skies (no wind) with a velocity (speed and direction) vector $\vec{v_1} = (400,100)$. Now, however, the air mass surrounding the plane is moving quickly ( $i.e.$, it's windy). Although Anjali has not adjusted any of the controls in the cockpit, the plane's new velocity is $\vec{v_2} = (350, 50)$. (Speeds are in kilometers per hour.) What is the speed of the wind?
Consider vector $\vec w$ (depicted below), which represents the wind. We can imagine how it would cause Anjali's airplane to slow down and change direction. It is reasonable to assume that the velocity of the wind added to the initial velocity of Anjali's airplane equals the resultant velocity of Anjali's airplane. $ \vec w + \vec{v_1} = \vec{v_2}$ We can now solve for $\vec w$. $\begin{aligned} \vec w + \vec{v_1} &= \vec{v_2}\\\\ \vec w &= \vec{v_2} - \vec{v_1}\\\\ \vec w &= (350,50) - (400,100)\\\\ \vec w &= (-50,-50) \end{aligned}$ We can find the magnitude of $\vec w$ (i.e., the speed of the wind) using the Pythagorean theorem. $\begin{aligned} \| \vec w \|^2 &= (-50)^2 + (-50)^2\\\\ \| \vec w \| &= \sqrt{2500 + 2500}\\\\ \| \vec w \| &= \sqrt{5000}\\\\ \| \vec w \| &\approx 70.7 \text{ km}/\text{h} \end{aligned}$ $\vec w$ is pointing into the third quadrant, so we can find its direction (call it $\theta~$ ) using the arctangent function and adding $\pi$. $\begin{aligned} \tan \theta &= \dfrac{-50}{-50}\\\\ \theta &= \arctan{(1) } \\\\ \theta &=\dfrac{\pi}{4} \end{aligned}$ Adding $\pi$ to this result gives us the actual direction, $\dfrac{5\pi}{4} \approx 3.93$ radians. The speed of the wind is $70.7$ kilometers per hour. The direction of the wind is $3.93$ radians.